At ${25}^{°}C$ and I atm pressure, the enthalpies of combustion are as given below:

Substance	$H_2$	C graphite	$C_2{H}_6\left(g\right)$
$\frac{{\Delta }_C{H}^0}{KJmo{l}^{-1}}$	-286.0	-394.0	-1560.0

 The enthalpy of formation of ethane is 

Given, $H_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}\to H_2{O}_{\left(\mathcal{l}\right)}$

$\Delta H°=-286.0KJ/mol$

$C\left(graphite\right)+O_{2\left(g\right)}\to C{O}_{2\left(g\right)};\Delta H°=-394.0KJmol$

$C_2{H}_{6\left(g\right)}+\frac{7}{2}{O}_{2\left(g\right)}\to 2C{O}_{2\left(g\right)}+3{H_{2O}}_{\left(\mathcal{l}\right)}$

$\Delta H°=-1560.0KJ/mol$

Enthalpy of formation of $C_2{H}_{6\left(g\right)}$  can be written as $2C\left(graphite\right)+3H_{2\left(g\right)}\to C_2{H}_{6\left(g\right)}$

So, by doing   $3\left(i\right)+2\left(ii\right)-\left(iii\right)$  we can get desired relation

$\Delta H_f^{°}\left(C_2{H}_6\right)=3x\left(-286\right)+2x\left(-394\right)-\left(-1560\right)$

$=-86.0kJmo{l}^{-1}$